A uniformly tapering conical wire is made from a material of young's modulus Y and has a normal unextended length `L` the radii at the upper and lower ends of this conical wire, have values R and 3R, respectively the upper end of the wire is fixed to a rigid support and a mass `M` is suspended from its lower end. the equilibrium extended length of this wire would equal to:

Options

A. a. `L(1+1/3(Mg)/(piYR^(2)))`
B. b. `L(1+2/3(Mg)/(piYR^(2)))`
C. c. `L(1+1/9(Mg)/(piYR^(2)))`
D. d. `L(1+2/9(Mg)/(piYR(2)))`

Explanation

To find the equilibrium extended length of a uniformly tapering conical wire under the influence of a suspended mass, we can follow these steps:

### Step 1: Understand the Geometry of the Wire
The conical wire has a radius \( R \) at the upper end and \( 3R \) at the lower end. The unextended length of the wire is \( L \).

### Step 2: Define the Variables
- Let \( Y \) be the Young's modulus of the material.
- Let \( M \) be the mass suspended from the lower end of the wire.
- The weight of the mass is given by \( W = Mg \), where \( g \) is the acceleration due to gravity.

### Step 3: Consider an Element of the Wire
Consider a small element of the wire at a distance \( x \) from the top, with a thickness \( dx \). The radius of this element can be expressed as a function of \( x \).

### Step 4: Establish the Relationship for Radius
Using similar triangles, the radius \( r \) at a distance \( x \) can be expressed as:
\[ r(x) = R + \frac{(3R - R)}{L} x = R + \frac{2R}{L} x = R(1 + \frac{2x}{L}) \]

### Step 5: Calculate the Area of Cross-Section
The area \( A \) of the cross-section of the wire at distance \( x \) is:
\[ A = \pi r^2 = \pi \left(R(1 + \frac{2x}{L})\right)^2 = \pi R^2 \left(1 + \frac{2x}{L}\right)^2 \]

### Step 6: Calculate the Stress and Strain
The stress \( \sigma \) on the element is given by:
\[ \sigma = \frac{W}{A} = \frac{Mg}{\pi R^2 \left(1 + \frac{2x}{L}\right)^2} \]

The strain \( \epsilon \) is given by:
\[ \epsilon = \frac{\sigma}{Y} = \frac{Mg}{Y \pi R^2 \left(1 + \frac{2x}{L}\right)^2} \]

### Step 7: Relate Strain to Elongation
The elongation \( dL \) of the small element \( dx \) can be expressed as:
\[ dL = \epsilon \cdot dx = \frac{Mg}{Y \pi R^2 \left(1 + \frac{2x}{L}\right)^2} dx \]

### Step 8: Integrate to Find Total Elongation
To find the total elongation \( \Delta L \) of the wire, integrate \( dL \) from \( 0 \) to \( L \):
\[ \Delta L = \int_0^L \frac{Mg}{Y \pi R^2 \left(1 + \frac{2x}{L}\right)^2} dx \]

### Step 9: Solve the Integral
Let \( u = 1 + \frac{2x}{L} \), then \( du = \frac{2}{L} dx \) or \( dx = \frac{L}{2} du \). The limits change from \( x = 0 \) to \( x = L \) which corresponds to \( u = 1 \) to \( u = 3 \):
\[ \Delta L = \int_1^3 \frac{Mg}{Y \pi R^2 u^2} \cdot \frac{L}{2} du = \frac{MgL}{2Y \pi R^2} \left[-\frac{1}{u}\right]_1^3 \]
\[ = \frac{MgL}{2Y \pi R^2} \left(-\frac{1}{3} + 1\right) = \frac{MgL}{2Y \pi R^2} \cdot \frac{2}{3} = \frac{MgL}{3Y \pi R^2} \]

### Step 10: Calculate the Total Extended Length
The total extended length \( L' \) of the wire is:
\[ L' = L + \Delta L = L + \frac{MgL}{3Y \pi R^2} \]
\[ = L \left(1 + \frac{Mg}{3Y \pi R^2}\right) \]

### Final Answer
The equilibrium extended length of the conical wire is:
\[ L' = L \left(1 + \frac{Mg}{3Y \pi R^2}\right) \]